# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: f1_0_main_ConstantStackPush, __init, f138_0_ack_GT
• Transitions: (pre-variables and post-variables)  f1_0_main_ConstantStackPush 1 f138_0_ack_GT: x1 = _arg1 ∧ x2 = _arg2 ∧ x1 = _arg1P ∧ x2 = _arg2P ∧ 10 = _arg2P ∧ 12 = _arg1P f138_0_ack_GT 2 f138_0_ack_GT: x1 = _x ∧ x2 = _x1 ∧ x1 = _x2 ∧ x2 = _x3 ∧ _x1 − 1 = _x3 ∧ 1 = _x2 ∧ _x ≤ 0 ∧ 0 ≤ _x1 − 1 f138_0_ack_GT 3 f138_0_ack_GT: x1 = _x4 ∧ x2 = _x5 ∧ x1 = _x6 ∧ x2 = _x7 ∧ _x5 = _x7 ∧ _x4 − 1 = _x6 ∧ 0 ≤ _x5 − 1 ∧ _x5 − 1 ≤ _x5 − 1 ∧ 0 ≤ _x4 − 1 f138_0_ack_GT 4 f138_0_ack_GT: x1 = _x8 ∧ x2 = _x9 ∧ x1 = _x10 ∧ x2 = _x11 ∧ _x9 − 1 = _x11 ∧ 0 ≤ _x9 − 1 ∧ _x9 − 1 ≤ _x9 − 1 ∧ 0 ≤ _x8 − 1 __init 5 f1_0_main_ConstantStackPush: x1 = _x12 ∧ x2 = _x13 ∧ x1 = _x14 ∧ x2 = _x15 ∧ 0 ≤ 0

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 f1_0_main_ConstantStackPush f1_0_main_ConstantStackPush f1_0_main_ConstantStackPush: x1 = x1 ∧ x2 = x2 __init __init __init: x1 = x1 ∧ x2 = x2 f138_0_ack_GT f138_0_ack_GT f138_0_ack_GT: x1 = x1 ∧ x2 = x2
and for every transition t, a duplicate t is considered.

### 2 SCC Decomposition

We consider subproblems for each of the 1 SCC(s) of the program graph.

### 2.1 SCC Subproblem 1/1

Here we consider the SCC { f138_0_ack_GT }.

### 2.1.1 Transition Removal

We remove transitions 2, 4 using the following ranking functions, which are bounded by 0.

 f138_0_ack_GT: x2

### 2.1.2 Transition Removal

We remove transition 3 using the following ranking functions, which are bounded by 0.

 f138_0_ack_GT: x1

### 2.1.3 Trivial Cooperation Program

There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.

## Tool configuration

AProVE

• version: AProVE Commit ID: unknown
• strategy: Statistics for single proof: 100.00 % (5 real / 0 unknown / 0 assumptions / 5 total proof steps)